I know that the formula for the angle of bending of light around the star [itex]\theta \;[/itex] in radians is:(adsbygoogle = window.adsbygoogle || []).push({});

[itex]\theta \; = \frac{4GM}{rc^2}[/itex]

This is equal to full bending of light, half of which occurs on the rear side of the star and the other half occuring on the foreground side.

So half of this is equal to:

[itex]\frac{\theta \;}{2} = \frac{2GM}{rc^2}[/itex]

So that:

[itex]dt = \frac{d\tau \;}{\sqrt{1-\frac{\theta\;}{2}}}[/itex]

[itex]\sqrt{1-\frac{\theta \;}{2}} = \frac{d\tau\;}{dt}[/itex]

[itex]1-\frac{\theta \;}{2} = \frac{d\tau\;^2}{dt^2}[/itex]

[itex]1-\frac{d\tau \;^2}{dt^2} = \frac{\theta\;}{2}[/itex]

[itex]2 \left( 1-\frac{d\tau \;^2}{dt^2} \right) = \theta\;[/itex]

If [itex]\theta \;=2 radians[/itex] then there would be infinite gravitational time dilation.

Did I get that right?

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# Bending of Light around a Star

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